ASE 376K
1.
Name: Philip L. Varghese
October 17, 2006
TEST #1
(Closed book section – max 30 minutes)
Explain why compressibility effects are negligible at low Mach numbers but very
significant at high Mach numbers.
(3 points)
dρ
du
For a compressible flow
= −M 2
. Because of the factor M2, the effect of velocity changes
ρ
u
on density is strongly attenuated at low Mach numbers (M<<1), but greatly amplified at high Mach
numbers (M>>1).
2.
Define the propulsive efficiency of a turbojet engine.
Thrust power generated
ηp =
=
Change in kinetic power of propellant stream
(words)
3.
T
(4 points)
ℑu
⎡
ue2 u 2 ⎤
m a ⎢ (1 + f ) − ⎥
2 2⎦
⎣
(symbols)
Sketch the T-s diagram for a non-ideal turbojet identifying all the relevant states a
(ambient), and inlet/exit from compressor, combustor, turbine, and nozzle. Draw
(6 points)
appropriate constant pressure lines and assume pe = pa.
To4 ≈ 2To3
o4
po4=rbpo3
ΔTot ≈ ΔToc
po2
o5
o3
To3
po5
po3
o6
po6
ΔToc
a
o2
1
2
3
4 5 6 7 (e)
pe = pa
p1
e
1
a
s
4.
Two turbojet engines have the following data. Engine A: TSFCA = 20 mg/s-N, Engine B:
TSFCB = 17 mg/s-N. Briefly explain which engine you would choose provided the costs
were comparable.
(2 points)
Engine B is more efficient because it has lower specific fuel consumption and would be cheaper to
operate. Hence assuming that both engines generate the required thrust and have comparable cost,
one would choose engine B.
5.
Sketch the variation of stagnation and static pressure along the typical flow streamline
shown passing through the non-ideal ramjet sketched below.
(5 points)
po,p
Note: Static pressure rises and stagnation pressure
falls across shocks
po
p
pa
x
6.
Sketch the variation of Thrust Specific Fuel Consumption vs. Specific Thrust for a turbojet
engine for variable maximum engine temperature and specified compressor pressure ratio
(πC). Show the variation for two different values of πC (e.g. (a) πC = 15 and (b) πC = 25)
and label the curves clearly. Show the low and high temperature ends of the lines on the
sketch.
(5 points)
Thigh
π c = 15
m f
ℑ
π c = 25
Tlow
ℑ / m a
______
ASE 376K
Test #1 Prob 1
Philip Varghese
Need: (a) ℑdesign , (b) At, (c) ηn for a rocket nozzle
pe
po, To
Given: m = 20 kg/s at sea-level, pa = 101 kPa,
Mˆ prop = 14 kg/kmol, γ = 1.18,
ue
t
ηpe = 0.95 (constant), po = 14 MPa, To = 3500 K
e
pa
Solution:
(1)
e + ( pe − pa )
ℑ = mu
(3)
0(2)
Assumptions:
(1) Steady, quasi-1D flow
(2) pe = pa at design point
(3) Adiabatic nozzle
(4) Ideal gas behavior, with
constant cp
Ae
(3)
Toe = To = 3500 K; Toe = Toi = Te +
ue2
2c p
η pe ( γ −1)
(2)
γ
η pe ( γ −1)
γ
⎛p ⎞
⎛p ⎞
From the equation sheet, for a polytropic expansion: Te = To ⎜ e ⎟
=1713 K
= To ⎜ a ⎟
⎝ po ⎠
⎝ po ⎠
(4)
(4)
8314 J / kmol − K
γ
Rˆ
R prop =
cp =
R prop ;
=
= 594 J/kg-K. So cp =3893 J/kg-K
ˆ
γ −1
14 kg / kmol
M prop
e = 74.6 kN = 75 kN
ue = 2c p (To − Te ) = 3.73×103 m/s and hence ℑ = mu
(3)
Flow is sonic at the throat, i.e., Mt = 1 and Tot = To . So Tt =
To
T
= o = 3211 K.
γ −1 2
γ +1
1+
Mt 1
2
2
γ
⎛ T ⎞η pe (γ −1)
= 7.72 MPa.
From the equation sheet, for a polytropic expansion: pt = po ⎜ t ⎟
⎝ To ⎠
(4)
pt
= 4.05 kg/m3.
ρt =
R propTt
ut = M t 1 γ RTt = 1.50 km/s (Equivalently: ut = 2c p (To − Tt ) = 1.50 km/s)
(1)
m = ρt At ut ⇒ At =
m
= 3.29×10–3 m2 = 32.9 cm2
ρt ut
⎛ p ⎞
From the isentropic relations: Tes = Toi ⎜ e ⎟
⎝ poi ⎠
T −T
ηn = o e = 0.97 (> ηpe as expected)
To − Tes
γ −1
γ
= 3500×0.471 = 1649 K
2
⎛u ⎞
(Equivalently: ues = 2c p (To − Tes ) = 3.80 km/s ; η n = ⎜ e ⎟ = 0.97)
⎝ ues ⎠
ASE 376K
Test #1 Prob 2
Philip Varghese
Need: m a , ηp for a ramjet
Given: Ma = 6 where Ta = 225 K, Tmax= To4 = 2700 K, ΔH R = 120 MJ/kg, ηb = 0.95, rb = 0.90,
rd = 0.85, rn = 0.92, γh = 1.28, Rair = 287 J/kg-K, ℑ = 60 kN
m f
u, Ma
ue
a
2
4
6 (e)
Solution:
(1)
ℑ = m a ⎡⎣(1 + f ) ue − u ⎤⎦ + ( pe − pa )
0(2)
Assumptions
(1) Steady, quasi-1D flow
(2) pe = pa at design point
(3) Ideal gas behavior,
Rair=287 J/kgK,
piecewise constant γc,
Ae
u = M a γ RTa = 1.80 km/s
⎛ γ −1 2 ⎞
To 2 = Toa = Ta ⎜1 + c
M a ⎟ = 1845 K,
2
⎝
⎠
(3)
c ph =
γh
γ h −1
γh
Rair = 1312 J/kg-K
(4) Adiabatic components
To 4 − To 2
(4)
Energy conservation equation applied to burner gives f =
⎡ηb ΔH R
⎤
− To 4 ⎥
⎢
⎢⎣ c ph
⎥⎦
γ
po 6 po 6 po 4 po 2 poa pa
=
pe
po 4 po 2 poa pa pe
1(2)
⎛ γ − 1 2 ⎞ γ −1
M a ⎟ = 0.704×1.58×103 = 1.11×103
= rn rb rd ⎜1 +
2
⎝
⎠
By assumption (4): To6 = To4 = 2700 K;
ue = 2c ph (To 6 − Te ) = 2.36 km/s
ℑ
= 104 kg/s
⎡⎣(1 + f ) ue − u ⎤⎦
ℑ⋅ u
ηp =
= 0.88
⎡
ue2 u 2 ⎤
m a ⎢(1 + f ) − ⎥
2 2⎦
⎣
m a =
= 1.02×10–2
⎛ p ⎞
Te = To 6 ⎜ e ⎟
⎝ po 6 ⎠
γ h −1
γh
= 582 K
ASE 376K
Test #1 Prob 3
Philip Varghese
Need: (a) ℑ (b) m f for a turbojet
m f , ab
m f ,b
Given: m a = 75 kg/s, Ma = 2
Ta = 216.5 K, pa = 7.56 kPa,
ηd = 0.86; πc = 15.5, ηc = 0.89,
rb = 0.94, ηb = 0.99, γh = 1.32,
ΔH R = 45 MJ/kg, To4 = 1525 K,
u
ue
2
a
ηt = 0.93, To6 = 1700 K, rab = 0.96,
ηn = 0.97
3
6
5
4
7 (e)
Solution:
(1)
ℑ = m a ⎡⎣(1 + f b + f ab ) ue − u ⎤⎦ + ( pe − pa )
0(2)
Ae
P γ
c pc = c Rair = 1005 J/kg-K; u = M a γ RTa = 590 m/s
γ c −1
(3)
u2
⎛ γ −1 2 ⎞
= 390 K)
Toa = Ta ⎜1 +
M ⎟ = 390 K, (OR To 2 = Toa = Ta +
2
2c pc
⎝
⎠
⎛ γ γ−1 ⎞
⎜ π −1 ⎟
ΔToc = To 2 ⎜ c
⎟ = 520 K, To 3 = To 2 + ΔToc = 910 K
⎜ ηc ⎟
⎝
⎠
Assumptions:
(1) Steady, quasi-1D flow
(2) pe = pa
(3) Ideal gas behavior,
Rair = 287 J/kg-K,
piecewise constant γc,
γh
(4) Adiabatic components
(5) No bleed air from
compressor, no
auxilliary power from
turbine
(4)
P γ
P
To 4 − To 3
c ph = h Rair = 1184 J/kg-K; fb =
= 1.70×10-2;
γ h −1
⎡ηb ΔH R
⎤
− To 4 ⎥
⎢
⎢⎣ c ph
⎥⎦
(3)
γh
(4,5)
ΔTot =
c pc
c ph (1 + fb )
ΔToc = 434 K.
To 5 = To 4 − ΔTot = 1091 K
The afterburner is on: To6 = 1700 K. f ab =
⎡ ΔT ⎤ γ h −1
π t = ⎢1 − ot ⎥ =0.2216
⎣ ηtTo 4 ⎦
m f ,ab (1 + f b )(To 6 − To 5 )
=1.78×10-2
=
ma
⎡ηab ΔH R
⎤
− To 6 ⎥
⎢
⎣⎢ c ph
⎦⎥
γ
po 6 po 6 po 5 po 4 po 3 po 2 pa
=
pe
po 5 po 4 po 3 po 2 pa pe
1(2)
γ − 1 2 ⎞ γ −1
⎛
= rabπ t rbπ c ⎜1 + ηd
M a ⎟ = 19.4
2
⎝
⎠
γ h −1
γh
⎛ p ⎞
= 1700×0.488 = 829 K;
ue = 2η n c ph (To 6 − Tes ) = 1.41×103 m/s
So Tes = To 6 ⎜ e ⎟
p
⎝ o6 ⎠
Finally, ℑ = m a ⎡⎣(1 + f b + f ab ) ue − u ⎤⎦ =6.55×104 N= 66 kN
m f = ( f b + f ab ) ⋅ m a = (1.70+1.78)×10-2 ×75kg/s = 2.61 kg/s; m f = 2.6 kg/s